How Fractions are Added and Subtracted

UNH Mathematics Center

1  How to do the additions

We add fractions of the same denomination (that is, having the same denominator) by adding or subtracting their numerators. The denominator remains, becoming the denominator of the combined fraction. For instance:

A G + B G + C G - D G - E G + F G = A+B+C-D-E+F G .

2  Why we need pre-addition conversions

Incidentally, you can also see why it's so convenient when numerical fractions are rewritten - if they can be, which isn't always the case - as decimals. When we add 0.342 + 1.036 = 1.378, the decimal point arrangement ensures that the fractions we are adding have the same denomination. The digits 2 and 6 in the last column represent fractions with the denominator 1000; when we add them ...

2 1000 + 6 1000 = 8 1000

as a part of doing the sum

0.34 2 + 1.03 6 = 1.37 8 ,

the '8' is also of the one-thousandth denomination, so it gets put in the third place to the right of the decimal point. The decimal point plan keeps track of the denominators for us. (It also keeps track of the denominators when we ''add and carry'' - but we'll speak of that some other time.)

3  How to do pre-addition conversions before adding

3.1  Multiplication by 1 is never frowned upon

If we wanted to, we could convert 3/(x²+2) so that its denominator is (x²+2)·cos(x) by multiplying by 1 in the form:

3 x2+2 · cos(x) cos(x) = 3·cos(x) (x2+2)cos(x)

3.2  Before converting, choose your new denomination prudently

So that we can convert each term according to our ''multiply by 1'' plan, our new denominator will need to be divisible by each of the individual denominators x, x³ and x²(x+1).

• We know it's possible. For instance, we could certainly do it by choosing as our new denominator the product of all three individual denominators: (x)(x³)(x²(x+1)) = x⁶(x+1).

• However, we know that if we were to do this, each and every one of our best friends would snicker sardonically behind his or her hands. Because there is a neater way!

Remember that the only requirement was that the new denominator be divisible by each individual denominator. So, let us look at the factors of individual denominators and see what we really need.

Each of these denominators has a x as a factor. The highest power of x that appears is x³, in the term

6y x3 .

So our new denominator must have x as a factor; but there is no reason why it needs to have a higher power than x³.

Also, one of the denominators has x+1 as a factor. This factor appears only in one of the terms, and only to the first power. But since it appears at all, our new denominator must include it.

There are no other factors in any of the terms. So our simplest (and therefore most satisfactory) new denominator will include

• THREE factors of x
• ONE factor x+1

A denominator all three terms can share - which is what we mean, literally, by a common denominator - is x³(x+1). Better and better, it is a simplest possible common denominator for these three terms. So, that is what we would use.

You probably remember doing this with numerical fractions in grade school, long ago. What we are looking for is not just a common denominator but a least common denominator.

3.3  Now make the conversions

Let's continue with our example

3 x + 6y x3 - 15 y-1 x2(x+1) .

Each of the fractional terms needs to be converted so that its new denominator is x³(x+1). And each conversion must have the effect (or non-effect?) of multiplying the term by 1:

• 3/x presently has x as its denominator, so we will want to multiply it by 1 in the form (x²(x+1))/(x²(x+1)). Rewritten, it will appear as
  

  3 x · x2(x+1) x2(x+1) = 3x2(x+1) x3(x+1)

• 6y/x³ presently has x³ as its denominator, so we will multiply it by 1 in the form (x+1)/(x+1). Then it will appear as
  

  6y x3 · x+1 x+1 = 6y(x+1) x3(x+1)

• 15[(y-1)/(x²(x+1))] has denominator x²(x+1) so we need only multiply it by x/x:
  

  15(y-1) x2(x+1) · x x = 15x(y-1) x3(x+1)

3.4  Now do the addition:

Now that the terms are all of the same denomination they can be added. We find:

3 x + 6y x3 - 15 y-1 x2(x+1) = 3x2(x+1) x3(x+1) + 6y(x+1) x3(x+1) - 15x(y-1) x3(x+1) = 3x2(x+1) + 6y(x+1) - 15x(y-1) x3(x+1)

3.4.1  How should we present the answer?

We would certainly expand the numerator of our sum to see if any terms can be combined:

3x2(x+1) + 6y(x+1) - 15x(y-1) x3(x+1) = 3x3+3x2+ 6xy +6y - 15xy +15x x3(x+1)

We see that two of the numerator's terms can can be added together, so that the numerator is more compactly written as 3x³+3x²-9xy+6y+15x.

Did you notice the sign of the last term in the numerator, 15x? It's correct. But the combination in this example: the minus sign before the original expression's third term, and the minus sign in its numerator y-1, is known to invite a particular error.

This amazing mistake - although it looks trivial, it's made astonishingly often - is to miss the use of the distributive rule when multiplying -15x by numerator, y-1. Specifically, the y gets properly multiplied by -15x, but the -1 does not. It's really a mistake in using the distributive rule when a fraction's numerator is a sum: because a number is multiplying a sum of numbers, the distributive rule applies. Perhaps the distributive rule is ''missed'' in this setting because no parentheses appeared aroung the numerator y-1 in the original example? We cannot say.

But, in this example you would strongly suspect that the dreaded non-distribution mistake had struck again, if the final term in the combined numerator were -15x instead of 15x.

Alas, this is not a random mistake - for many students, this one error happens over and over again. Worse, each and every time it happens, it gets passed by as a ''little minus sign error.'' Make sure it doesn't happen to you.

We could leave the fraction with its numerator all expanded into separate terms - although in this example we'd probably take notice of the common factor 3, and make the minimal effort of rewriting that. But we would leave the denominator factored!

3(x3+x2-3xy+2y+5x) x3(x+1)

We have two reasons for not multiplying out the denominator:

1. Sometimes (not here, admittedly) it's just too much like work.
2. The fraction's expression is more informative if we leave the denominator ''as is.'' Left factored, we can see that the denominator is zero if x = 0 or x = -1. We can even see what the sign of the denominator is: for instance, if -1 < x < 0, x³ is negative and x+1 is positive, which would make the denominator negative. These things would not be nearly so easy to spot if we multiplied the denominator out.

4  One more example, with square roots:

It happens that the derivative of the function

(2x+1)   ___ Öx-3

is

2   ___ Öx-3   +     2x+1 2   ___ Öx-3

Depending on when you are reading this ... it could be, of course, that you don't know that derivative yet. But, to be ready when you do know about derivatives, you should be able to simplify sums like this - because you'll want to know where they might be zero, or positive, or negative.

We begin by selecting a simplest common denominator. The individual terms' denominators are 1, and the square-root expression. So we will use

2   ___ Öx-3

as a common denominator.

We make the conversion:

• The first term's denominator is 1; to convert it, we multiply it by 1 in a convenient guise:
  

  2   ___ Öx-3   ·   ___ Öx-3   ___ Öx-3 = 4(x-3) 2   ___ Öx-3

• The second term's denominator is already suitable, so we do not change it at all.

Short pause here: Do you know the fact about square roots that allowed us to eliminate the radical in the numerator? A number is the square of its square root: 5 = (Ö5)². So, when we multiply a square root by itself, what we get is ''the thing under the square root,'' that is to say the radicand. In this case, the square of

___ Öx-3

is the radicand itself, x-3.

After converting the individual terms so that they share the same denominator, we add. We get

4(x-3)  +  2x+1 2   ___ Öx-3   = 6x - 11 2   ___ Öx-3

We can tell quite a bit about this expression's values, just by looking at it.

• It has no value at all unless x > 3; if x = 3 the denominator is zero, and if x < 3 the square root makes no sense - in the world of the real numbers, where we are living during this calculus course.
• The numerator would be zero if x = 11/6. But that won't happen, because x must be at least 3 for the entire fraction to make sense.
• When x > 3, the numerator is positive - in fact its value is at least 7, because 6x is at least 18. And square roots are always positive. (They are defined that way so that the square root can be used as a function, which must always produce a single specific output number for each of its possible input numbers.)

Putting these bits and pieces together, we can see - after we've added the fractions in this expression - that its values are always positive.

It was not nearly so easy to see this from the appearance of the expression as it first arrived on our doorstep. The reason for this is that we first saw this expression as

2   ___ Öx-3   +     2x+1 2   ___ Öx-3

which is a sum of terms. By adding these (fractional) terms, we converted it to to

6x - 11 2   ___ Öx-3

so that we can look instead at one numerator, and one denominator. This way we can use our knowledge of how real numbers act when they are multiplied and divided to see what the expression's sign might be - and when it might be zero, or without a value.

5  Problems for You to Work

1. Here are some terms to be added and subtracted. In each case select a simplest common denominator. Present your sum in an appropriately simple form.
   1. 
      

      4 x-2 - 3 x-1 - 5 x+1 (x-1)(x-2)

   2. 
      

      t+1 tu + 5 t2u - u-1 u2

   3. 
      

      5 y2-2y - y y2-4

2. For each of the expressions below, first add the fractions. By looking at the factors of the sum fraction, decide which values of the variable make the sum positive. (You'll recall that a fraction's value is positive, if its numerator and denominator have the same sign.)
   1. 
      

      6 - 2x2+1 x2

   2. 
      

      y (y+5)2 - y2 (y+5)3

   3. 
      

      2r·   ___ Ör+3    +  r2 2   ___ Ör+3